MHB Ryan's question at Yahoo Answers (Eigenvalues of A^*A)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
AI Thread Summary
All eigenvalues of the matrix A*A are proven to be nonnegative by showing that for any eigenvalue λ, it can be expressed as the ratio of the squared norm of Ax to the squared norm of x, which is always nonnegative. Additionally, the matrix A*A + I is shown to be invertible by demonstrating that if -1 were an eigenvalue of A*A, it would lead to a contradiction regarding the determinant. Thus, the determinant of A*A + I is non-zero, confirming its invertibility. This discussion provides a clear understanding of these linear algebra concepts. Further inquiries can be directed to the specified math help forum.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Here is the question:

Let A be a mxn matrix. Prove that all eigenvalues of A*A are nonnegative and prove that A*A + I is invertible.

THANK YOU!

Here is a link to the question:

Linear Algebra Proof Help Please? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello Ryan,

Suppose $\lambda$ is an eigenvalue of $A^*A\in \mathbb{C}^{n\times n}$ then, there exists $x=(x_1,\ldots,x_n)^T\in\mathbb{C}^n$ such that $A^*Ax=\lambda x$. Multypling both sides by $x^*=(\overline{x_1},\ldots,\overline{x_n})$ we get $$x^*A^*Ax=(Ax)^*(Ax)=\lambda x^*x$$ But $y=Ax=(y_1,\ldots,y_m)^T$ and $x\neq 0$ (i.e. $||x||\neq0$), so $$\lambda=\frac{y^*y}{x^*x}=\frac{||y||^2}{||x||^2}\ge 0$$ On the other hand, $$\det (A^*A+I)=0\Leftrightarrow \det (A^*A-(-1)I)=0\Leftrightarrow -1\mbox{ is eigenvalue of }A^*A$$ This is a contradiction, so $\det(A^*A+I)\neq0$, as a consequence $A^*A+I$ is invertible.

If you have further questions, you can post them in the http://www.mathhelpboards.com/f14/ section.
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top